Problem: Find $\dfrac{d}{dx}\left[\ln(x)\cdot e^{2x}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $e^{ 2x}\left(\dfrac1x+2\ln(x)\right)$ (Choice B) B $\dfrac{e^{ 2x}}{x}$ (Choice C) C $\dfrac{2e^{ 2x}}{x}$ (Choice D) D $e^{ 2x}\left(\dfrac1x+\ln(x)\right)$
Explanation: $\ln(x)\cdot e^{2x}$ is a product of a function and a composite function. Let... $u(x)=\ln(x)$ $v(x)=e^x$ $w(x)=2x$... then $\ln(x)\cdot e^{2x}=u(x)\cdot v\Bigl(w(x)\Bigr)$. To find $\dfrac{d}{dx}\left[\ln(x)\cdot e^{2x}\right]$, we will need to use the product rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v\Bigl(w(x)\Bigr)\right] \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \gray{\text{Product rule}} \\\\ &=u'(x)\cdot v\Bigl(w(x)\Bigr)+u(x)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\dfrac{1}{x}$ $v'(x)=e^x$ $w'(x)=2$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot{ v\Bigl(w(x)\Bigr)}+u(x)\cdot{ v'\Bigl(w(x)\Bigr)}\cdot w'(x) \\\\ &=\dfrac{1}{x}\cdot{e^{2x}}+\ln(x)\cdot{e^{ 2x}}\cdot2 \\\\ &=e^{ 2x}\left(\dfrac1x+2\ln(x)\right) \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[\ln(x)\cdot e^{2x}\right]= e^{ 2x}\left(\dfrac1x+2\ln(x)\right)$.